How Do You Differentiate Double Integrals with Non-Differentiable Functions?

[jrsh]

differentiating double integrals -- help please!

Hello,

Could somebody please help me with my problem? I have a double
integral

F(t) = Integral from 0 to t of f(s,t). w.r.t. d(g(s))
f(s,t) = Integral from t to A of h(s,u). w.r.t. d u.

Where "g(s)" is a function of "s", which may or may not be
differentiable w.r.t. s. In this case, how can I differentiate "F(x)"
w.r.t. "x", or calculate "d(F(x))"?

I'd really appreciate your help.

Thanks,

 

[HallsofIvy]

Quick answer is that the derivative with respect to "x" is 0 because there is no "x" in your formula!

I assume you mean with respect to t.

If g is not differentiable, then F will not have a derivative at points at which g is not differentiable so we may as well assume g is differentiable.

\frac{d}{dt}\[\int_0^t\int_t^A h(s,u)du g'(s)ds

Looking only at the outer integral, the derivative with respect to t, by the Fundamental Theorem of Calculus, is f(t,t)g'(t) where g is differentiable and does not exist where g is not differentiable. From the inner integral,
f(t,t)= \int_t^A h(t,u)du
Therefore, the derivative is
g'(t)\int_t^A h(t,u)du

 

[jrsh]

Thanks!

In fact, I'm struggling with the case where g isn't differentiable. I think I'm troubled by a fundamental problem: if g isn't differentiable, then what is

d(\int_0^t f(x) d(g(x)))

Here, I'm not differentiating

\int_0^t f(x) d(g(x))

w.r.t. t, but calculating the change of it if there's a small increment of t. In this case, what is the result? Following the fundamental theorem of calculus, and if I modify the theorem a bit, it seems I can get

f(t) d(g(t))

However, I have no idea whether the above result can be derived.

Further, both the fundamental theorem of calculus and Leibniz rule require we differentiate an integral and the integral is integrating some function w.r.t a variable. Why is there no "generalized" version where the integral is w.r.t some function that may be non-differentiable?

Thanks a lot! I truly look forward to your reply (have been struggling with this and have got really frustrated)